Integrand size = 28, antiderivative size = 187 \[ \int \frac {(a+i a \tan (c+d x))^4}{(e \sec (c+d x))^{15/2}} \, dx=\frac {2 a^4 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {e \sec (c+d x)}}{33 d e^8}+\frac {2 a^4 \sin (c+d x)}{55 d e^5 (e \sec (c+d x))^{5/2}}+\frac {2 a^4 \sin (c+d x)}{33 d e^7 \sqrt {e \sec (c+d x)}}-\frac {4 i a (a+i a \tan (c+d x))^3}{15 d (e \sec (c+d x))^{15/2}}-\frac {4 i \left (a^4+i a^4 \tan (c+d x)\right )}{55 d e^2 (e \sec (c+d x))^{11/2}} \] Output:
2/33*a^4*cos(d*x+c)^(1/2)*InverseJacobiAM(1/2*d*x+1/2*c,2^(1/2))*(e*sec(d* x+c))^(1/2)/d/e^8+2/55*a^4*sin(d*x+c)/d/e^5/(e*sec(d*x+c))^(5/2)+2/33*a^4* sin(d*x+c)/d/e^7/(e*sec(d*x+c))^(1/2)-4/15*I*a*(a+I*a*tan(d*x+c))^3/d/(e*s ec(d*x+c))^(15/2)-4/55*I*(a^4+I*a^4*tan(d*x+c))/d/e^2/(e*sec(d*x+c))^(11/2 )
Time = 3.00 (sec) , antiderivative size = 155, normalized size of antiderivative = 0.83 \[ \int \frac {(a+i a \tan (c+d x))^4}{(e \sec (c+d x))^{15/2}} \, dx=-\frac {i a^4 \sqrt {e \sec (c+d x)} \left (64+112 \cos (2 (c+d x))+48 \cos (4 (c+d x))-54 i \sin (2 (c+d x))-37 i \sin (4 (c+d x))+40 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) (i \cos (4 (c+d x))+\sin (4 (c+d x)))\right ) (\cos (4 (c+2 d x))+i \sin (4 (c+2 d x)))}{660 d e^8 (\cos (d x)+i \sin (d x))^4} \] Input:
Integrate[(a + I*a*Tan[c + d*x])^4/(e*Sec[c + d*x])^(15/2),x]
Output:
((-1/660*I)*a^4*Sqrt[e*Sec[c + d*x]]*(64 + 112*Cos[2*(c + d*x)] + 48*Cos[4 *(c + d*x)] - (54*I)*Sin[2*(c + d*x)] - (37*I)*Sin[4*(c + d*x)] + 40*Sqrt[ Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*(I*Cos[4*(c + d*x)] + Sin[4*(c + d *x)]))*(Cos[4*(c + 2*d*x)] + I*Sin[4*(c + 2*d*x)]))/(d*e^8*(Cos[d*x] + I*S in[d*x])^4)
Time = 0.87 (sec) , antiderivative size = 205, normalized size of antiderivative = 1.10, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {3042, 3977, 3042, 3977, 3042, 4256, 3042, 4256, 3042, 4258, 3042, 3120}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \frac {(a+i a \tan (c+d x))^4}{(e \sec (c+d x))^{15/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(a+i a \tan (c+d x))^4}{(e \sec (c+d x))^{15/2}}dx\) |
| \(\Big \downarrow \) 3977 |
| \(\displaystyle \frac {a^2 \int \frac {(i \tan (c+d x) a+a)^2}{(e \sec (c+d x))^{11/2}}dx}{5 e^2}-\frac {4 i a (a+i a \tan (c+d x))^3}{15 d (e \sec (c+d x))^{15/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {a^2 \int \frac {(i \tan (c+d x) a+a)^2}{(e \sec (c+d x))^{11/2}}dx}{5 e^2}-\frac {4 i a (a+i a \tan (c+d x))^3}{15 d (e \sec (c+d x))^{15/2}}\) |
| \(\Big \downarrow \) 3977 |
| \(\displaystyle \frac {a^2 \left (\frac {7 a^2 \int \frac {1}{(e \sec (c+d x))^{7/2}}dx}{11 e^2}-\frac {4 i \left (a^2+i a^2 \tan (c+d x)\right )}{11 d (e \sec (c+d x))^{11/2}}\right )}{5 e^2}-\frac {4 i a (a+i a \tan (c+d x))^3}{15 d (e \sec (c+d x))^{15/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {a^2 \left (\frac {7 a^2 \int \frac {1}{\left (e \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{7/2}}dx}{11 e^2}-\frac {4 i \left (a^2+i a^2 \tan (c+d x)\right )}{11 d (e \sec (c+d x))^{11/2}}\right )}{5 e^2}-\frac {4 i a (a+i a \tan (c+d x))^3}{15 d (e \sec (c+d x))^{15/2}}\) |
| \(\Big \downarrow \) 4256 |
| \(\displaystyle \frac {a^2 \left (\frac {7 a^2 \left (\frac {5 \int \frac {1}{(e \sec (c+d x))^{3/2}}dx}{7 e^2}+\frac {2 \sin (c+d x)}{7 d e (e \sec (c+d x))^{5/2}}\right )}{11 e^2}-\frac {4 i \left (a^2+i a^2 \tan (c+d x)\right )}{11 d (e \sec (c+d x))^{11/2}}\right )}{5 e^2}-\frac {4 i a (a+i a \tan (c+d x))^3}{15 d (e \sec (c+d x))^{15/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {a^2 \left (\frac {7 a^2 \left (\frac {5 \int \frac {1}{\left (e \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2}}dx}{7 e^2}+\frac {2 \sin (c+d x)}{7 d e (e \sec (c+d x))^{5/2}}\right )}{11 e^2}-\frac {4 i \left (a^2+i a^2 \tan (c+d x)\right )}{11 d (e \sec (c+d x))^{11/2}}\right )}{5 e^2}-\frac {4 i a (a+i a \tan (c+d x))^3}{15 d (e \sec (c+d x))^{15/2}}\) |
| \(\Big \downarrow \) 4256 |
| \(\displaystyle \frac {a^2 \left (\frac {7 a^2 \left (\frac {5 \left (\frac {\int \sqrt {e \sec (c+d x)}dx}{3 e^2}+\frac {2 \sin (c+d x)}{3 d e \sqrt {e \sec (c+d x)}}\right )}{7 e^2}+\frac {2 \sin (c+d x)}{7 d e (e \sec (c+d x))^{5/2}}\right )}{11 e^2}-\frac {4 i \left (a^2+i a^2 \tan (c+d x)\right )}{11 d (e \sec (c+d x))^{11/2}}\right )}{5 e^2}-\frac {4 i a (a+i a \tan (c+d x))^3}{15 d (e \sec (c+d x))^{15/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {a^2 \left (\frac {7 a^2 \left (\frac {5 \left (\frac {\int \sqrt {e \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{3 e^2}+\frac {2 \sin (c+d x)}{3 d e \sqrt {e \sec (c+d x)}}\right )}{7 e^2}+\frac {2 \sin (c+d x)}{7 d e (e \sec (c+d x))^{5/2}}\right )}{11 e^2}-\frac {4 i \left (a^2+i a^2 \tan (c+d x)\right )}{11 d (e \sec (c+d x))^{11/2}}\right )}{5 e^2}-\frac {4 i a (a+i a \tan (c+d x))^3}{15 d (e \sec (c+d x))^{15/2}}\) |
| \(\Big \downarrow \) 4258 |
| \(\displaystyle \frac {a^2 \left (\frac {7 a^2 \left (\frac {5 \left (\frac {\sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)}}dx}{3 e^2}+\frac {2 \sin (c+d x)}{3 d e \sqrt {e \sec (c+d x)}}\right )}{7 e^2}+\frac {2 \sin (c+d x)}{7 d e (e \sec (c+d x))^{5/2}}\right )}{11 e^2}-\frac {4 i \left (a^2+i a^2 \tan (c+d x)\right )}{11 d (e \sec (c+d x))^{11/2}}\right )}{5 e^2}-\frac {4 i a (a+i a \tan (c+d x))^3}{15 d (e \sec (c+d x))^{15/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {a^2 \left (\frac {7 a^2 \left (\frac {5 \left (\frac {\sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{3 e^2}+\frac {2 \sin (c+d x)}{3 d e \sqrt {e \sec (c+d x)}}\right )}{7 e^2}+\frac {2 \sin (c+d x)}{7 d e (e \sec (c+d x))^{5/2}}\right )}{11 e^2}-\frac {4 i \left (a^2+i a^2 \tan (c+d x)\right )}{11 d (e \sec (c+d x))^{11/2}}\right )}{5 e^2}-\frac {4 i a (a+i a \tan (c+d x))^3}{15 d (e \sec (c+d x))^{15/2}}\) |
| \(\Big \downarrow \) 3120 |
| \(\displaystyle \frac {a^2 \left (\frac {7 a^2 \left (\frac {5 \left (\frac {2 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {e \sec (c+d x)}}{3 d e^2}+\frac {2 \sin (c+d x)}{3 d e \sqrt {e \sec (c+d x)}}\right )}{7 e^2}+\frac {2 \sin (c+d x)}{7 d e (e \sec (c+d x))^{5/2}}\right )}{11 e^2}-\frac {4 i \left (a^2+i a^2 \tan (c+d x)\right )}{11 d (e \sec (c+d x))^{11/2}}\right )}{5 e^2}-\frac {4 i a (a+i a \tan (c+d x))^3}{15 d (e \sec (c+d x))^{15/2}}\) |
Input:
Int[(a + I*a*Tan[c + d*x])^4/(e*Sec[c + d*x])^(15/2),x]
Output:
(((-4*I)/15)*a*(a + I*a*Tan[c + d*x])^3)/(d*(e*Sec[c + d*x])^(15/2)) + (a^ 2*((7*a^2*((2*Sin[c + d*x])/(7*d*e*(e*Sec[c + d*x])^(5/2)) + (5*((2*Sqrt[C os[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[e*Sec[c + d*x]])/(3*d*e^2) + ( 2*Sin[c + d*x])/(3*d*e*Sqrt[e*Sec[c + d*x]])))/(7*e^2)))/(11*e^2) - (((4*I )/11)*(a^2 + I*a^2*Tan[c + d*x]))/(d*(e*Sec[c + d*x])^(11/2))))/(5*e^2)
Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2 )*(c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x _)])^(n_), x_Symbol] :> Simp[2*b*(d*Sec[e + f*x])^m*((a + b*Tan[e + f*x])^( n - 1)/(f*m)), x] - Simp[b^2*((m + 2*n - 2)/(d^2*m)) Int[(d*Sec[e + f*x]) ^(m + 2)*(a + b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 + b^2, 0] && GtQ[n, 1] && ((IGtQ[n/2, 0] && ILtQ[m - 1/2, 0]) || (EqQ[n, 2] && LtQ[m, 0]) || (LeQ[m, -1] && GtQ[m + n, 0]) || (ILtQ[m, 0] & & LtQ[m/2 + n - 1, 0] && IntegerQ[n]) || (EqQ[n, 3/2] && EqQ[m, -2^(-1)])) && IntegerQ[2*m]
Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[Cos[c + d*x]*(( b*Csc[c + d*x])^(n + 1)/(b*d*n)), x] + Simp[(n + 1)/(b^2*n) Int[(b*Csc[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1] && IntegerQ[2* n]
Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(b*Csc[c + d*x] )^n*Sin[c + d*x]^n Int[1/Sin[c + d*x]^n, x], x] /; FreeQ[{b, c, d}, x] && EqQ[n^2, 1/4]
Time = 46.42 (sec) , antiderivative size = 151, normalized size of antiderivative = 0.81
| method | result | size |
| default | \(\frac {a^{4} \left (\frac {2 \sin \left (d x +c \right ) \left (88 \cos \left (d x +c \right )^{6}-16 \cos \left (d x +c \right )^{4}+3 \cos \left (d x +c \right )^{2}+5\right )}{165}+\frac {2 i \left (-88 \cos \left (d x +c \right )^{7}+60 \cos \left (d x +c \right )^{5}\right )}{165}+\frac {2 i \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {EllipticF}\left (i \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ), i\right ) \left (5+5 \sec \left (d x +c \right )\right )}{165}\right )}{d \sqrt {e \sec \left (d x +c \right )}\, e^{7}}\) | \(151\) |
| risch | \(-\frac {i {\mathrm e}^{i \left (d x +c \right )} \left (11 \,{\mathrm e}^{6 i \left (d x +c \right )}+47 \,{\mathrm e}^{4 i \left (d x +c \right )}+81 \,{\mathrm e}^{2 i \left (d x +c \right )}+85\right ) a^{4} \sqrt {2}}{1320 d \,e^{7} \sqrt {\frac {e \,{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}}+\frac {2 \sqrt {-i \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}\, \sqrt {i \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}\, \sqrt {i {\mathrm e}^{i \left (d x +c \right )}}\, \operatorname {EllipticF}\left (\sqrt {-i \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}, \frac {\sqrt {2}}{2}\right ) a^{4} \sqrt {e \,{\mathrm e}^{i \left (d x +c \right )} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}}{33 d \sqrt {e \,{\mathrm e}^{3 i \left (d x +c \right )}+e \,{\mathrm e}^{i \left (d x +c \right )}}\, e^{7} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) \sqrt {\frac {e \,{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}}\) | \(273\) |
| parts | \(\frac {a^{4} \left (\frac {2 \sin \left (d x +c \right ) \left (77 \cos \left (d x +c \right )^{6}+91 \cos \left (d x +c \right )^{4}+117 \cos \left (d x +c \right )^{2}+195\right )}{1155}+\frac {2 i \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {EllipticF}\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right ) \left (-195-195 \sec \left (d x +c \right )\right )}{1155}\right )}{d \sqrt {e \sec \left (d x +c \right )}\, e^{7}}+\frac {a^{4} \left (\frac {2 \sin \left (d x +c \right ) \left (77 \cos \left (d x +c \right )^{6}-119 \cos \left (d x +c \right )^{4}+12 \cos \left (d x +c \right )^{2}+20\right )}{1155}+\frac {2 i \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {EllipticF}\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right ) \left (-20-20 \sec \left (d x +c \right )\right )}{1155}\right )}{d \sqrt {e \sec \left (d x +c \right )}\, e^{7}}-\frac {4 i a^{4} \left (\frac {2 \cos \left (d x +c \right )^{7}}{15}-\frac {2 \cos \left (d x +c \right )^{5}}{11}\right )}{d \,e^{7} \sqrt {e \sec \left (d x +c \right )}}-\frac {8 i a^{4}}{15 d \left (e \sec \left (d x +c \right )\right )^{\frac {15}{2}}}-\frac {6 a^{4} \left (\frac {2 \sin \left (d x +c \right ) \left (-77 \cos \left (d x +c \right )^{6}+14 \cos \left (d x +c \right )^{4}+18 \cos \left (d x +c \right )^{2}+30\right )}{1155}+\frac {2 i \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {EllipticF}\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right ) \left (-30-30 \sec \left (d x +c \right )\right )}{1155}\right )}{d \sqrt {e \sec \left (d x +c \right )}\, e^{7}}\) | \(443\) |
Input:
int((a+I*a*tan(d*x+c))^4/(e*sec(d*x+c))^(15/2),x,method=_RETURNVERBOSE)
Output:
a^4/d*(2/165*sin(d*x+c)*(88*cos(d*x+c)^6-16*cos(d*x+c)^4+3*cos(d*x+c)^2+5) +2/165*I*(-88*cos(d*x+c)^7+60*cos(d*x+c)^5)+2/165*I*(cos(d*x+c)/(cos(d*x+c )+1))^(1/2)*EllipticF(I*(cot(d*x+c)-csc(d*x+c)),I)*(1/(cos(d*x+c)+1))^(1/2 )*(5+5*sec(d*x+c)))/(e*sec(d*x+c))^(1/2)/e^7
Time = 0.10 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.66 \[ \int \frac {(a+i a \tan (c+d x))^4}{(e \sec (c+d x))^{15/2}} \, dx=\frac {-80 i \, \sqrt {2} a^{4} \sqrt {e} {\rm weierstrassPInverse}\left (-4, 0, e^{\left (i \, d x + i \, c\right )}\right ) + \sqrt {2} {\left (-11 i \, a^{4} e^{\left (8 i \, d x + 8 i \, c\right )} - 58 i \, a^{4} e^{\left (6 i \, d x + 6 i \, c\right )} - 128 i \, a^{4} e^{\left (4 i \, d x + 4 i \, c\right )} - 166 i \, a^{4} e^{\left (2 i \, d x + 2 i \, c\right )} - 85 i \, a^{4}\right )} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c\right )}}{1320 \, d e^{8}} \] Input:
integrate((a+I*a*tan(d*x+c))^4/(e*sec(d*x+c))^(15/2),x, algorithm="fricas" )
Output:
1/1320*(-80*I*sqrt(2)*a^4*sqrt(e)*weierstrassPInverse(-4, 0, e^(I*d*x + I* c)) + sqrt(2)*(-11*I*a^4*e^(8*I*d*x + 8*I*c) - 58*I*a^4*e^(6*I*d*x + 6*I*c ) - 128*I*a^4*e^(4*I*d*x + 4*I*c) - 166*I*a^4*e^(2*I*d*x + 2*I*c) - 85*I*a ^4)*sqrt(e/(e^(2*I*d*x + 2*I*c) + 1))*e^(1/2*I*d*x + 1/2*I*c))/(d*e^8)
Timed out. \[ \int \frac {(a+i a \tan (c+d x))^4}{(e \sec (c+d x))^{15/2}} \, dx=\text {Timed out} \] Input:
integrate((a+I*a*tan(d*x+c))**4/(e*sec(d*x+c))**(15/2),x)
Output:
Timed out
\[ \int \frac {(a+i a \tan (c+d x))^4}{(e \sec (c+d x))^{15/2}} \, dx=\int { \frac {{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{4}}{\left (e \sec \left (d x + c\right )\right )^{\frac {15}{2}}} \,d x } \] Input:
integrate((a+I*a*tan(d*x+c))^4/(e*sec(d*x+c))^(15/2),x, algorithm="maxima" )
Output:
integrate((I*a*tan(d*x + c) + a)^4/(e*sec(d*x + c))^(15/2), x)
\[ \int \frac {(a+i a \tan (c+d x))^4}{(e \sec (c+d x))^{15/2}} \, dx=\int { \frac {{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{4}}{\left (e \sec \left (d x + c\right )\right )^{\frac {15}{2}}} \,d x } \] Input:
integrate((a+I*a*tan(d*x+c))^4/(e*sec(d*x+c))^(15/2),x, algorithm="giac")
Output:
integrate((I*a*tan(d*x + c) + a)^4/(e*sec(d*x + c))^(15/2), x)
Timed out. \[ \int \frac {(a+i a \tan (c+d x))^4}{(e \sec (c+d x))^{15/2}} \, dx=\int \frac {{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^4}{{\left (\frac {e}{\cos \left (c+d\,x\right )}\right )}^{15/2}} \,d x \] Input:
int((a + a*tan(c + d*x)*1i)^4/(e/cos(c + d*x))^(15/2),x)
Output:
int((a + a*tan(c + d*x)*1i)^4/(e/cos(c + d*x))^(15/2), x)
\[ \int \frac {(a+i a \tan (c+d x))^4}{(e \sec (c+d x))^{15/2}} \, dx=\frac {\sqrt {e}\, a^{4} \left (\int \frac {\sqrt {\sec \left (d x +c \right )}}{\sec \left (d x +c \right )^{8}}d x +\int \frac {\sqrt {\sec \left (d x +c \right )}\, \tan \left (d x +c \right )^{4}}{\sec \left (d x +c \right )^{8}}d x -4 \left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \tan \left (d x +c \right )^{3}}{\sec \left (d x +c \right )^{8}}d x \right ) i -6 \left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \tan \left (d x +c \right )^{2}}{\sec \left (d x +c \right )^{8}}d x \right )+4 \left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \tan \left (d x +c \right )}{\sec \left (d x +c \right )^{8}}d x \right ) i \right )}{e^{8}} \] Input:
int((a+I*a*tan(d*x+c))^4/(e*sec(d*x+c))^(15/2),x)
Output:
(sqrt(e)*a**4*(int(sqrt(sec(c + d*x))/sec(c + d*x)**8,x) + int((sqrt(sec(c + d*x))*tan(c + d*x)**4)/sec(c + d*x)**8,x) - 4*int((sqrt(sec(c + d*x))*t an(c + d*x)**3)/sec(c + d*x)**8,x)*i - 6*int((sqrt(sec(c + d*x))*tan(c + d *x)**2)/sec(c + d*x)**8,x) + 4*int((sqrt(sec(c + d*x))*tan(c + d*x))/sec(c + d*x)**8,x)*i))/e**8